additivity of arc length I was looking over a proof that every rectifiable curve is parametrizable by its arc length and found that it was used that the arc length is additive.
It is pretty obvious that this is true, but how can you show it rigorously?
For a curve $c: \mathbb R \supset [a,b] \rightarrow (X,d)$ in a metric space $(X,d)$ the arc length is defined as
$$L(c_{|_{[a,b]}})=\sup \left\{\sum _{i=1}^{n}d(c(t_{i}),c(t_{i-1})):n\in \mathbb {N} {\text{ and }}a=t_{0}<t_{1}<\dotsb <t_{n}=b\right\}.$$
For any $u < v < w \in [a,b]$ how can I show that $L(c_{|_{[u,w]}})=L(c_{|_{[u,v]}}) + L(c_{|_{[v,w]}})$ ?
 A: There exists some $t^*$ such that $c(t^*) = v$
For any partition $t_0<t_1<\cdots<t_k\le t^*<t_{k+1}\cdots<t_n$ 
$L(c_{|u,w|}) = \sup \{\sum_\limits {i=1}^k d(c(t_i), c(t_{i-1})+\sum_\limits {i=k+1}^n d(c(t_i), c(t_{i-1})\} = L(c_{|u,v|}) + L(c_{|v,w|})$
A: $If\space C_{[a,b]},\space C_{[a,c]},\space and\space C_{[c,b]}\space are\space the\space curves\space parametrized\space by\space g(t)\space for\space t\in[a,b],\space t\in[a,c],\space and\space t\in[c,b],\space claim\space L(C_{[a,b]})=L(C_{[a,c]})+L(C_{[c,b]}).$
For a partition $P=\{a=t_{0}<t_{1}<\dotsb <t_{n}=b\}$ of $[a,b]$, defined $L_g(P)=\sum_{i=1}^{n}||g(t_i)-g(t_{i-1})||.$
The arc lengths of $C_{[a,b]}$, $C_{[a,c]}$, and $C_{[c,b]}$ are
$\begin{aligned}&L(C_{[a,b]})=sup\{L_g(P):P\space is\space a\space partition\space of\space[a,b]\},\space \\&L(C_{[a,c]})=sup\{L_g(P):P\space is\space a\space partition\space of\space [a,c]\},\space\\and\space&L(C_{[b,c]})=sup\{L_g(P):P\space is\space a\space partition\space of\space[b,c]\}.\end{aligned}$.
By Def. of sup,
$\begin{aligned}\forall\epsilon>0, &\exists\space a\space partition\space P_{[a,b]}\space of\space[a,b]\space where\space c\in P_{[a,b]}\ni L_{[a,b]}-L_g(P_{[a,b]})<\epsilon,\\&\exists\space a\space partition\space Q_{[a,c]}\space of\space[a,c]\ni L_{[a,c]}-L_g(Q_{[a,c]})<\frac\epsilon2,\\&\exists\space a\space partition\space Q_{[c,b]}\space of\space[c,b]\ni L_{[c,b]}-L_g(Q_{[c,b]})<\frac\epsilon2.\end{aligned}$
Then,
$\begin{aligned}L(C_{[a,b]})-\epsilon<L_g(P_{[a,b]})&=L_g(P_{[a,b]}\cap[a,c])+L_g(P_{[a,b]}\cap[c,b])\\&\le L(C_{[a,c]})+L(C_{[c,b]}).\end{aligned}$
Moreover,
$\begin{aligned} L(C_{[a,c]})+L(C_{[c,b]})&<L_g(Q_{[a,c]})+L_g(Q_{[c,b]})+\frac\epsilon2+\frac\epsilon2\\&=L_g(Q_{[a,c]}\cup Q_{[c,b]})+\epsilon\\&\le L(C_{[a,b]})+\epsilon.\end{aligned}$
Thus,
$L(C_{[a,b]})-\epsilon<L(C_{[a,c]})+L(C_{[c,b]})<L(C_{[a,b]})+\epsilon.$
Since we can choose $\epsilon>0$ to be arbitrary small, we have
$L(C_{[a,b]})=L(C_{[a,c]})+L(C_{[c,b]}).$
